A1128. N Queens Puzzle
The "eight queens puzzle" is the problem of placing eight chess queens on an 8×8 chessboard so that no two queens threaten each other. Thus, a solution requires that no two queens share the same row, column, or diagonal. The eight queens puzzle is an example of the more general N queens problem of placing N non-attacking queens on an N×N chessboard. (From Wikipedia - "Eight queens puzzle".)
Here you are NOT asked to solve the puzzles. Instead, you are supposed to judge whether or not a given configuration of the chessboard is a solution. To simplify the representation of a chessboard, let us assume that no two queens will be placed in the same column. Then a configuration can be represented by a simple integer sequence (Q1, Q2, ..., QN), where Qi is the row number of the queen in the i-th column. For example, Figure 1 can be represented by (4, 6, 8, 2, 7, 1, 3, 5) and it is indeed a solution to the 8 queens puzzle; while Figure 2 can be represented by (4, 6, 7, 2, 8, 1, 9, 5, 3) and is NOT a 9 queens' solution.
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Figure 1 | Figure 2 |
Input Specification:
Each input file contains several test cases. The first line gives an integer K (1 < K <= 200). Then K lines follow, each gives a configuration in the format "N Q1 Q2 ... QN", where 4 <= N <= 1000 and it is guaranteed that 1 <= Qi <= N for all i=1, ..., N. The numbers are separated by spaces.
Output Specification:
For each configuration, if it is a solution to the N queens problem, print "YES" in a line; or "NO" if not.
Sample Input:
4
8 4 6 8 2 7 1 3 5
9 4 6 7 2 8 1 9 5 3
6 1 5 2 6 4 3
5 1 3 5 2 4
Sample Output:
YES
NO
NO
YES
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
int num[], N, K, hashTB[];
int main(){
scanf("%d", &K);
for(int i = ; i < K; i++){
scanf("%d", &N);
fill(hashTB, hashTB + , );
for(int j = ; j <= N; j++){
scanf("%d", &num[j]);
hashTB[num[j]]++;
}
int tag = ;
for(int j = ; j <= N; j++){
if(hashTB[j] != ){
tag = ;
break;
}
int m = j - , n = num[j] - ;
while(m >= && m <= N && j >= && j <= N){
if(num[m] == n){
tag = ;
break;
}
m--; n--;
}
m = j + ; n = num[j] + ;
while(m >= && m <= N && j >= && j <= N){
if(num[m] == n){
tag = ;
break;
}
m++; n++;
}
}
if(tag == )
printf("NO\n");
else printf("YES\n");
}
cin >> N;
return ;
}
总结:
1、由于已经保证了不在同一列,所以只需要检查行和斜线即可。
2、检查a、b两点间的斜线,可用abs(Xa - Xb) == abs(Ya - Yb)。
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